By Proposition REF , it suffices to consider the \(\operatorname{L\acute{e}vy}(0,1)\)  distribution. Let
\(f\)  be its density function, i.e., \(f(x)\equiv f(x;0,1)\)  as in (REF ).
Fix \(p\in \lbrace 0\rbrace \) . To show that \(\nu (\cdot )\)  is increasing at \(p\) , it suffices to
show that the log density ratio of the left and right parts
\(R_p(x) := \log \left(\frac{f(\nu _p-x)}{f(\nu _p+x)}\right)= \frac{3}{2}(\log (\nu _p + x)-\log (\nu _p - x)) - \frac{1}{2(\nu _p - x)} + \frac{1}{2(\nu _p + x)},\) 
for \(x\in [0,\nu _p)\) , has exactly one positive critical point.
Indeed, observe that \(R_p(x)\rightarrow -\infty \)  as \(x\rightarrow \nu _p-\)  and that \(R_p(0)=0\) . Moreover,
(REF ) implies that \(R_p(\cdot )\)  cannot be non-positive for all \(x>0\) .
Therefore, if \(R_p(\cdot )\)  has a single positive critical point, then it must be a maximum,
and it follows that the conditions of Lemma REF  are satisfied and so
\(\nu (\cdot )\)  is increasing at \(p\) .
To identify the positive critical points, observe that
\(R^{\prime }_p(x) = \frac{3}{2(\nu _p - x)} + \frac{3}{2(\nu _p + x)} - \frac{1}{2(\nu _p - x)^2}- \frac{1}{2(\nu _p + x)^2},\) 
so the critical points are solutions to the equation
\(3(\nu _p-x)(\nu _p+x)^2 + 3(\nu _p+x)(\nu _p-x)^2 - (\nu _p+x)^2 - (\nu _p - x)^2 = 0.\) 
Further simplification yields a quadratic equation in \(x\) ,
\((3\nu _p+2)x^2 + \nu _p^2(3\nu _p-2) = 0\) 
which has exactly one root in \((0,\nu _p)\)  if \(\nu _p>2/3\) .
This holds for arbitrary \(p\in , so we may concludetrue positive skewness if \) 1>2/3\( and \) 1>0\( hold, by Lemma \ref {nup clopen}.The mode and median of the Lévy distribution are well-known (see, e.g., \cite {Nolan20})and can be computed directly as \) 0=1/3\( and \) 1=12(1-erfc(12))-2
2.17\(, which completes the proof.\)
Fix \(k\in \mathbb {N}\)  and let \(f\)  be the density function of \(\chi ^2(k)\) , i.e., \(f(x)\equiv f(x;k)\)  as in (REF ).
Fix \(p\in \lbrace 0\rbrace \) . There are two cases.
Case 1: \(k=1,2\) . Here, \(f(x)\)  is decreasing on its domain and so true positive skewness follows from
Proposition REF  and Theorem REF .
Case 2: \(k>2\) . As in the proof of Theorem REF , to show that \(\nu (\cdot )\)  is increasing at \(p\) , it
suffices to show that the ratio of the left and right parts
\(R_p(x) := \log \left(\frac{f(\nu _p-x)}{f(\nu _p+x)}\right)= \left(\frac{k}{2} - 1\right) \log \left(\frac{\nu _p-x}{\nu _p+x}\right)+x\) 
for \(x\in [0,\nu _p)\) , has exactly one positive critical point, since \(R_p(x)\rightarrow -\infty \)  as \(x\rightarrow \nu _p-\)  and \(R_p(0)=0\) .
Observe that
\(R^{\prime }_p(x) = \frac{(2-k)\nu _p}{\nu _p^2-x^2}+1,\) 
so the critical points are solutions to the equation \(x^2 - \nu _p^2 + (k-2)\nu _p = 0\) , i.e.,
\(x = \pm \sqrt{\nu _p^2-(k-2)\nu _p}.\) 
There is exactly one positive critical point when \(\nu _p^2-(k-2)\nu _p>0\) . Since \(f\)  has support on the positive half
line, then \(\nu _p\)  is always non-negative by Proposition REF . Thus \(\nu (\cdot )\)  is increasing at \(p\)  if
\(\nu _p-k+2>0\) . Now by Lemma REF , true positive skewness of \(\chi ^2(k)\)  follows if the median satisfies
\(\nu _1>k-2\) , but this inequality is well-known (see, e.g., Sen [1]} and the references therein).
By Proposition REF , it suffices to consider the \(\operatorname{Weibull}(k,1)\)  distribution.
Let \(f\)  be its density function, i.e., \(f(x)\equiv f(x;k,1)\)  as in (REF ).
It is well-known (see, e.g., [2]}) that the \(\operatorname{Weibull}(k,1)\)  distribution
has finite moments of all orders and is unimodal with median and mode given by
\(\nu _0 = \left(\frac{k-1}{k}\right)^{1/k},\qquad \nu _1 = (\log 2)^{1/k}.\) 
Notice that \(\nu _0<\nu _1\)  holds if and only if \(k<(1-\log 2)^{-1}\) .
Since the Weibull distribution has support on the positive half-line, then the second
part of the theorem follows from Proposition REF . For the first part
of the theorem, the “only if” part follows from (REF ).
It remains to show the “if” part. For \(k\le 1\) , \(f\)  is strictly decreasing, so we are
done by Proposition REF .
Suppose \(k>1\)  and fix \(p\ge 1\) . As in the proof of Theorem REF , to show that
\(\nu (\cdot )\)  is increasing at \(p\) , it suffices to show that the log density ratio of the
left and right parts
\(R_p(x):=\log \left(\frac{f(\nu _p-x)}{f(\nu _p+x)}\right)= (k-1)\left[\log (\nu _p-x)-\log (\nu _p+x)\right]-(\nu _p-x)^k+(\nu _p+x)^k\) 
for \(x\in [0,\nu _p)\) , has exactly one positive critical point, since \(R_p(x)\rightarrow -\infty \)  as
\(x\rightarrow \nu _p-\)  and \(R_p(0)=0\) . Observe that
\(R^{\prime }_p(x) = k(\nu _p-x)^{k-1}+k(\nu _p+x)^{k-1}-\frac{k-1}{\nu _p-x}-\frac{k-1}{\nu _p+x},\) 
so the critical points are solutions to the equation
\(k(\nu _p^2-x^2)\left[(\nu _p-x)^{k-1}+(\nu _p+x)^{k-1}\right]-2(k-1)\nu _p = 0.\) 
Since \(x\in [0,\nu _p)\) , the binomial series for \((\nu _p-x)^{k-1}\)  and
\((\nu _p+x)^{k-1}\)  converge, hence
\((\nu _p-x)^{k-1}+(\nu _p+x)^{k-1} = 2\sum _{n=0}^\infty \binom{k-1}{2n} \nu _p^{k-2n-1} x^{2n},\) 
in the notation of the generalized binomial coefficient. Substituting into (REF )
yields the equation
\(g(x) := 2k\sum _{n=1}^\infty \left(\binom{k-1}{2n} - \binom{k-1}{2n-2}\right)\nu _p^{k-2n+1} x^{2n} + 2\nu _p(k\nu _p^k-k+1) = 0,\) 
where
\(\binom{k-1}{2n}-\binom{k-1}{2n-2} = \frac{k(k-1)\dots (k-2n+2)(k+1-4n)}{(2n)!}.\) 
We analyze the sign changes of the coefficients in (REF ) to determine the number
of its positive roots by splitting into several cases for the value of \(k\) .
Case 1: \(2<k<3\) . Then \(k,k-1,k-2\)  are positive and \(k-3,\dots ,k-2n+2\)  are negative.
There are an even number (possibly zero, if \(n=1,2\) ) of negative factors in \(k(k-1)\dots (k-2n+2)\) ,
so it is positive. Also note that \(k+1-4n<0\)  for all \(n\ge 1\) , hence (REF ) is
negative for all \(n\ge 1\) .
For the series expression \(g(x)=\sum _{m=0}^\infty a_mx^m\) , we have shown that \(a_m<0\)  holds for
non-zero even \(m\)  and \(a_m=0\)  holds for odd \(m\) . By Descartes' rule of signs for infinite series,
\(g^*_p\)  has no positive real roots if \(a_0\le 0\)  and has at most one positive real root if \(a_0>0\) .
Suppose \(a_0\le 0\)  such that \(g\)  has no positive real roots. By extension, \(R^{\prime }_p\)  has no positive
real roots, so \(R_p\)  has no positive extrema and is strictly monotonic on \([0,\nu _p)\) . Since
\(g_p(0)=0\)  and \(\lim _{x\rightarrow \nu _p-}g_p(x)=-\infty \) , then \(g_p\)  is strictly negative on \([0,\nu _p)\) ,
hence \(f(\nu _p-x)<f(\nu _p+x)\)  for all \(x\in (0,\nu _p)\) . By monotonicity of the integral, this
contradicts (REF ). Thus \(a_0>0\)  holds, and \(g\)  has exactly one positive root
in \((0,\nu _p)\) , which implies that \(\nu (\cdot )\)  is increasing at \(p\) .
This holds for
arbitrary \(p\ge 1\) , so \(\nu (\cdot )\)  is increasing on \(\lbrace 0\rbrace \) . Moreover, since \(a_0=2\nu _p(k\nu _p^k-k+1)>0\)  holds, then \(\nu _p>\left(\frac{k-1}{k}\right)^{1/k}=\nu _0\)  holds for every
\(p\ge 1\) , and we are done.
Case 2: \(1<k<2\) . The inequality \(k+1-4n<0\)  still holds for all \(n\ge 1\) , but we now have
\(k,k-1>0\)  and \(k-2,\dots ,k-2n+2<0\) . If \(n>1\) , then there are an odd number
of negative factors in \(k(k-1)\dots (k-2n+2)\) , so the numerator of (REF ) is positive.
If \(n=1\) , then the numerator is simply \(k(k-3)<0\) .
Thus our coefficients in \(g(x)=\sum _{m=0}^\infty a_mx^m\)  are positive for even \(m>2\)  and zero for
odd \(m\) , with \(a_2<0\) . By Descartes' rule of signs, \(g\)  has at most two positive real roots if
\(a_0>0\)  and at most one if \(a_0\le 0\) .
Suppose \(a_0\le 0\) . If \(g\)  has a single positive root, then \(\lim _{x\rightarrow \nu _p-}g(x)>0\) . But
by continuity this limit tends to \(-2(k-1)\nu _p<0\) . If instead \(g^*_p\)  has no positive real roots,
then by the same argument above we contradict (REF ). Thus \(a_0>0\) . In this
case, if \(g\)  has zero or two positive roots, then again \(\lim _{x\rightarrow \nu _p-}g(x)>0\)  and we reach a
contradiction. Hence \(g\)  has exactly one positive real root. We may now conclude in the same
fashion as in Case 1.
Case 3: \(3<k<\frac{1}{1-\log 2}\) . We again look at the numerator of (REF ). If
\(n=1\) , the numerator is \(k(k-3)>0\) . If \(n=2\) , the numerator is \(k(k-1)(k-2)(k-7)<0\) . If \(n>2\) , then
\(k(k-1)\dots (k-2n+2)\)  has positive factors \(k,\dots ,k-3\) , and an odd number of negative factors
\(k-4,\dots ,k-2n+2\) . Additionally, \(k+1-4n<0\)  when \(n>2\) , so the numerator of (REF ) is
positive.
Our coefficients in \(g(x)=\sum _{m=0}^\infty a_mx^m\)  are zero for odd \(m\) , positive for even
\(m\ge 6\) , negative for \(m=4\) , and positive for \(m=2\) . This yields two sign changes and hence at most
two positive roots of \(g\)  if \(a_0>0\) . Recall from Case 1 that \(a_0>0\)  holds if and only if
\(\nu _p>\nu _0\)  holds. If the latter holds, then \(g\)  must have exactly one positive root since \(g\) 
is negative at the right limit of its domain, and so \(\nu (\cdot )\)  is increasing at \(p\) . Now by
Lemma REF , true positive skewness follows if \(\nu _1>\nu _0\) , but this holds
immediately from (REF ) and our assumption \(k<\frac{1}{1-\log 2}\) .
Case 4: \(k=2\) . We can plug \(k=2\)  directly into (REF ) to obtain the roots
\(x^2 = \nu _p^2-\frac{1}{2} = \nu _p^2-\nu _0^2.\) 
The argument in Case 1 can be adapted to show that \(\nu _p>\nu _0\)  holds for every \(p\ge 1\) ,
hence exactly one positive root exists.
Case 5: \(k=3\) . Similarly, we plug \(k=3\)  into (REF ) and obtain the roots
\(x^4 = \nu _p^4-\frac{2}{3}\nu _p = \nu _p(\nu _p^3-\nu _0^3)\) 
Again, since \(\nu _p>\nu _0\)  holds for every \(p\ge 1\) , then exactly one positive root exists.
Recall that \(\phi ,\Phi \)  are the density and distribution functions of the standard Gaussian
distribution. For simplicity, set
\(\Psi (x) := \sqrt{2\pi }\,\Phi (x) = \int _{-\infty }^xe^{-t^2/2}\,dt.\) 
We show true positive skewness for positive shape parameter \(\alpha >0\) ; the proof for true
negative skewness for \(\alpha <0\)  is analogous, and clearly the skew-normal is simply the normal
distribution, which is symmetric, when \(\alpha =0\) . Thus, fix \(\alpha >0\)  and let \(f\)  be the
density function of \(\operatorname{SkewNormal}(\alpha )\) , i.e., \(f(x)\equiv f(x;\alpha )\)  as in
(REF ).
Fix \(p\in \lbrace 0\rbrace \) . To show that \(\nu (\cdot )\)  is increasing at \(p\) , it suffices to show
that the log density ratio of the left and right parts
\(R_p(x) := \log \left(\frac{f(\nu _p-x)}{f(\nu _p+x)}\right)= 2\nu _px + \log \Psi (\alpha (\nu _p-x)) - \log \Psi (\alpha (\nu _p+x))\) 
has exactly one positive root \(x_0\) , satisfying \(R_p(x)>0\)  for \(0<x<x_0\)  and \(R_p(x)<0\)  for
\(x>x_0\) . This condition holds if the following four conditions can be verified:

\(R_p(0) = 0\) ;

\(R^{\prime \prime }_p(x)<0\)  for all \(x>0\) .

\(R^{\prime }_p(0)>0\) ; and

\(R^{\prime }_p(x)<0\)  for some \(x>0\) .

Condition (I) holds trivially. To prove condition (II), define
\(\psi (x) := \log \Psi (x),\) 
hence
\(R_p(x) &= 2\nu _px + \psi (\alpha (\nu _p-x)) - \psi (\alpha (\nu _p+x)); \nonumber \\R_p^{\prime }(x) &= 2\nu _p - \alpha \psi ^{\prime }(\alpha (\nu _p-x)) - \alpha \psi ^{\prime }(\alpha (\nu _p+x)); \\R_p^{\prime \prime }(x) &= \alpha ^2\psi ^{\prime \prime }(\alpha (\nu _p-x)) - \alpha ^2\psi ^{\prime \prime }(\alpha (\nu _p+x)).\nonumber \) 

Condition (II) holds if and only if \(\psi ^{\prime \prime }(\alpha (\nu _p-x)) < \psi ^{\prime \prime }(\alpha (\nu _p+x))\)  holds
for every \(x>0\) . It suffices then to show that \(\psi ^{\prime \prime }\)  is monotonically increasing everywhere,
i.e., that \(\psi ^{\prime \prime \prime }(x)>0\)  holds for every \(x\in \mathbb {R}\) .
The third logarithmic derivative of \(\Psi \)  is given by
\(\psi ^{\prime \prime \prime }(x) &= \frac{\Psi ^{\prime \prime \prime }(x)}{\Psi (x)} - \frac{3\Psi ^{\prime }(x)\Psi ^{\prime \prime }(x)}{\Psi (x)^2}+ 2\left(\frac{\Psi ^{\prime }(x)}{\Psi (x)}\right)^3 \nonumber \\&= \frac{x^2-1}{\Psi (x)}\,e^{-x^2/2} + \frac{3x}{\Psi (x)^2}\,e^{-x^2}+ \frac{2}{\Psi (x)^3}\,e^{-3x^2/2} \nonumber \\&= \frac{e^{-3x^2/2}}{\Psi (x)^3}\Big [(x^2-1)e^{x^2}\Psi (x)^2 + 3xe^{x^2/2}\Psi (x) + 2\Big ].\) 

One may compute directly
\(\begin{split}\psi ^{\prime \prime \prime }(1) &= \frac{e^{-3/2}}{\Psi (1)^3}\Big [3e^{1/2}\Psi (1) + 2\Big ] > 0; \\\psi ^{\prime \prime \prime }(-1) &= \frac{e^{-3/2}}{\Psi (-1)^3}\Big [-3e^{1/2}\Psi (-1)+2\Big ] \approx 0.12 > 0,\end{split}\) 
so if \(\psi ^{\prime \prime \prime }\)  has no zeroes, then continuity implies that \(\psi ^{\prime \prime \prime }\)  is positive everywhere.
Define \(u\equiv u(x):=(\psi ^{\prime }(x))^{-1}=e^{x^2/2}\Psi (x)\) , noticing that \(u\)  is strictly
positive. By (REF ), \(\psi ^{\prime \prime \prime }(x)=0\)  holds if and only if
\((x^2-1)u^2 + 3xu + 2 = 0,\) 
which, by the quadratic formula, holds exactly when
\(u = \frac{-3x \pm \sqrt{x^2+8}}{2(x^2-1)} = \mp \frac{4}{\sqrt{x^2+8}\pm 3x} =: Q^\pm (x).\) 
In other words, the zeroes of \(h^{\prime \prime \prime }\)  are solutions to the equations
\(u(x) = Q^\pm (x),\) 
with special care needed for the asymptotic values \(x=\pm 1\)  for \(Q^\pm \) . We handled these
cases in (REF ), so it suffices to consider \(x\ne \pm 1\) .
Observe that \(Q^\pm (x)<0<u(x)\)  holds if \(x>1\) , so solutions to (REF ) must fall in
\((-\infty ,1)\) . If \(x<-1\) , then both \(Q^+(x)\)  and \(Q^-(x)\)  are positive and, one can easily
verify the inequalities \(0<Q^-(x)<Q^+(x)\) . On the other hand, if \(-1<x<1\) , then \(Q^+(x)<0<u(x)\) 
whereas \(Q^-(x)>0\) . Therefore, it suffices to show that \(u(x) < Q^-(x)\)  for every \(x<1\) ,
\(x\ne -1\) .
The key observation is that \(u(x)\)  is simply the reflection of the Mills' ratio of the standard
Gaussian, i.e.,
\(u(-x) = M(x) := \frac{1-\Phi (x)}{\phi (x)} = e^{x^2/2}\int _x^\infty e^{-t^2/2}\,dt,\) 
for which very precise bounds in terms of elementary functions exist. In particular, we use
the known bound (see, e.g., [3]}, or [4]}, [5]})
\(M(x) < \frac{4}{\sqrt{x^2+8}+3x},\qquad \forall \,x>-1,\) 
which yields
\(u(x) < \frac{4}{\sqrt{x^2+8}-3x} = Q^-(x),\qquad \forall \,x<1,\,x\ne -1.\) 
We have now shown that (REF ) has no solutions and thus that \(\psi ^{\prime \prime \prime }\)  has no zeroes,
proving condition (II).
Since \(R_p(\cdot )\)  cannot be non-positive on the entire half-line, as a consequence of
(REF ), then condition (II) implies condition (III). For condition (IV),
observe that \(\psi ^{\prime }(x)=e^{-x^2/2}\,\Psi (x)^{-1}\)  is decreasing and tends to infinity as
\(x\rightarrow -\infty \) , so (REF ) implies that \(R^{\prime }_p(x)\rightarrow -\infty \)  as \(x\rightarrow \infty \) .
It remains to show that \(\nu _p>\nu _0\)  for every \(p\in \lbrace 0\rbrace \) . Notice that
\(R^{\prime }_p(0)=2\nu _p-2\alpha \psi ^{\prime }(\alpha \nu _p)\)  is increasing as a function of \(\nu _p\)  given
the decrease of \(\psi ^{\prime }\) . The mode \(\nu _0\)  is the unique value satisfying \(f^{\prime }(\nu _0)=0\) , which
is equivalent to the equality \(R^{\prime }_0(0)=2\nu _0-2\alpha \psi ^{\prime }(\alpha \nu _0)=0\) . Condition (III)
and the increase of \(\nu _p\mapsto R^{\prime }_p(0)\)  now yield \(\nu _0<\nu _p\)  for every
\(p\in \lbrace 0\rbrace \) .

Proofs of results for Section  REF
To prove the first part of the theorem, fix \(p\in X\)  and define
\(f_n(a):=E|X_n-a|^p,\quad a\in \mathbb {R}; \quad \quad \quad \nu _n:=\operatornamewithlimits{arg\,min}_{a\in \mathbb {R}}f_n(a);\) 
\(f(a):=E|X-a|^p,\quad a\in \mathbb {R}; \quad \quad \quad \nu :=\operatornamewithlimits{arg\,min}_{a\in \mathbb {R}}f(a);\) 
so that \(\nu _n\)  and \(\nu \)  are the \(p\) -means of \(X_n\)  and \(X\)  respectively by Definition
REF .
Let \(a_n\)  be any sequence of numbers converging to \(a\) .
Observe that the random variable \(X_n-a_n\)  has distribution function \(\tilde{F}_n(x)=F_n(x+a_n)\) .
Similarly, \(X-a\)  has distribution function \(\tilde{F}(x)=F(x+a)\) , so the uniform convergence
\(F_n\rightarrow F\)  implies the pointwise convergence \(\tilde{F}_n(x)\rightarrow \tilde{F}(x)\) . Therefore, \(X_n-a_n\) 
converges to \(X-a\)  in distribution.
By Jensen's inequality, we have
\(\sup _nE|X_n-a_n|^{p+\epsilon } \le 2^{p+\epsilon }\sup _n(E|X_n|^{p+\epsilon }+ |a_n|^{p+\epsilon })<\infty ,\) 
where finiteness holds by the convergence of \(a_n\rightarrow a\)  and by assumption on the \(X_n\) 's.
It follows from, e.g., [6]}, that
\(f_n(a_n)=E|X_n-a_n|^p\rightarrow E|X-a|^p=f(a).\) 
Suppose that the \(\nu _n\) 's are contained in a compact interval. Then every subsequence
\(\nu _{n_k}\)  has a limit point \(\nu _*\) . Since \(f_{n_k}(\nu _{n_k})\le f_{n_k}(a)\)  for all \(a\in \mathbb {R}\) ,
then by taking \(k\rightarrow \infty \)  we obtain \(f(\nu _*)\le f(a)\)  holds for every \(a\in \mathbb {R}\)  via
(REF ). Then \(\nu _*\)  minimizes \(f\) , and since the minimizer of \(f\)  is unique
(see Definition REF ), we obtain \(\nu _*=\nu \) . Every subsequence of \(\nu _n\)  converges
to \(\nu \) , hence \(\nu _n\rightarrow \nu \) .
Consider \(\nu _n\)  and \(\nu \)  as functions of \(p\) , so \(\nu _n\)  converges pointwise to \(\nu \) . True
non-negative skewness of \(X_n\)  implies \(\nu _n\)  is non-decreasing, and the pointwise limit of
monotone functions is monotone, hence \(\nu \)  is non-decreasing.
It remains to show that the \(\nu _n\) 's are contained in a compact interval. Suppose otherwise,
so we have a subsequence \(\nu _{n_k}\rightarrow \infty \)  as \(k\rightarrow \infty \) . (The argument is similar if
\(\nu _{n_k}\rightarrow -\infty \) .) We have from (REF ) the pointwise convergence \(f_n\rightarrow f\) ,
so for fixed \(\delta >0\) , we may choose large enough \(K\)  such that \(\nu _{n_k}>\nu \)  and \(|f_{n_k}(\nu )-f(\nu )|<\delta \)  for all \(k\ge K\) . Since \(f_{n_k}\)  is strictly convex with minimizer
\(\nu _{n_k}\) , then \(f_{n_k}(a)<f(\nu )+\delta \)  for all \(a\in [\nu ,\nu _{n_k}]\) .
Clearly \(f\rightarrow \infty \)  as \(a\rightarrow \infty \) , so choose \(x>\nu _{n_K}>\nu \)  large enough such that
\(f(x)>f(\nu )+2\delta \) . For any \(N>0\) , there exists \(k>K\)  large enough such that
\(n_k>N\)  and \(\nu _{n_k}>x\)  such that \(f_{n_k}(x)<f(\nu )+\delta <f(x)-\delta \)  as shown above.
This contradicts the pointwise convergence \(f_{n_k}(x)\rightarrow f(x)\) , and so concludes the proof of
the first part.
The second part of the theorem now easily follows from the fact that \(\nu _n(p)\rightarrow \nu _n(p)\) .
Indeed, for \(p>p^{\prime }\) , we have by assumption that \(\nu _n(p)-\nu _n(p^{\prime })>C(p-p^{\prime })\)  holds for every
\(n\in \mathbb {N}\) , so taking limits yields \(\nu (p)-\nu (p^{\prime })\ge C(p-p^{\prime })>0\) , hence \(\nu (\cdot )\)  is strictly
increasing.

Proofs of results for Section  REF
Let \(X\)  be a continuous random variable with density function \(f_X\)  decreasing on its support \((0, R)\)  for possibly infinite \(R\) . Let \(Y=u(X)\)  where \(u\)  is convex and strictly increasing on the support of \(X\) . We define \(w=u^{-1}\) . Note that \(X=Y=[1,\infty )\) . We write \( for this domain. Let \) fY\( be the density of \) Y\(. It suffices to show that\begin{equation}\log \left(\frac{f_Y(\nu _p - y)}{f_Y(\nu _p + y)}\right)\end{equation}changes sign exactly once on the interval \) (0, )\(, for all \) p. First note that as
\(y\)  increases, \(\nu _p-y\)  approaches 0. Once \(y>\nu _p\) , \(f_Y(\nu _p - y)= 0\) . Thus, on the
interval \(y\in (\nu _p, \infty )\) ,
\(\log \left(\frac{f_Y(\nu _p - y)}{f_Y(\nu _p + y)}\right) = -\infty \) 
On the interval \(u\in (0, \nu _p]\) , we have \(\lbrace \nu _p - y\rbrace  \in [0, \nu _p)\) , and \(\lbrace \nu _p + y\rbrace \in [\nu _p, 2\nu _p)\) . We expand \(f_Y\) :
\(\log \left(\frac{f_Y(\nu _p - y)}{f_Y(\nu _p + y)}\right)&= \log \left(\frac{-f_X[w(\nu _p - y)]w^{\prime }(\nu _p - y)}{-f_X[w(\nu _p + y)]w^{\prime }(\nu _p + y)}\right)\nonumber \\&=\log \left(\frac{f_X[w(\nu _p - y)]}{f_X[w(\nu _p + y)]}\right) +\log \left(\frac{w^{\prime }(\nu _p - y)}{w^{\prime }(\nu _p + y)}\right)\) 

Because \(w=u^{-1}\)  is an increasing function and \(f_X\)  is a decreasing function,
\(f_X[w(\nu _p - y)] > f_X[w(\nu _p + y)]\) 
\(\log \left(\frac{f_X[w(\nu _p - y)]}{f_X[w(\nu _p + y)]}\right) > 0\) 
The function \(u\)  is convex and increasing, \(w\)  is concave, and therefore \(w^{\prime }\)  is decreasing.
It follows that
\(w^{\prime }(\nu _p - y) > w^{\prime }(\nu _p + y)\) 
\(\log \left(\frac{w^{\prime }(\nu _p - y)}{w^{\prime }(\nu _p + y)}\right) > 0\) 
which implies that (REF ) is strictly positive on \((0, \nu _p]\) . Since ()
is strictly positive on \((0, \nu _p]\)  and strictly negative on \((\nu _p, \infty )\) , it changes
sign exactly once on the interval \((0, \infty )\) . This satisfies Lemma REF  for
all \(p\in , so we obtain true positive skewness.\)
Let \(f\)  denote the density function of \(X\) . For \(p\in \lbrace 0\rbrace \) , define
\(h_p(c):=f(\nu _p+c)-f(\nu _p-c),\ c\in [0,\nu _p]\) 
and
\(\mathcal {S}_p:=\lbrace c>0:h_p(c)>0\rbrace .\) 
If \(\mathcal {S}_p\)  is non-empty, then its infimum
\(c_p:=\inf \mathcal {S}_p\) 
exists and is non-negative. Note that if \(f\)  is continuous, then so is \(h_p\) . Then \(\mathcal {S}_p\) 
is the preimage of an open set under \(h_p\) , so \(\mathcal {S}_p\)  is also open and \(c_p\notin \mathcal {S}_p\) .
Since \(h_p(0)=0\) , then continuity implies
\(h_p(c_p)=0.\) 
Similarly, if \(f\)  is differentiable, then so is \(h_p\) . Because \(h_p(c)\le 0\)  for all \(c<c_p\) , then
\(h_p^{\prime }(c_p)\ge 0.\) 
To prove the theorem, we will require several lemmas.
Lemma 3.1 
The density \(f\)  is convex on \((0,\theta _1)\cup (\theta _2,\infty )\) , concave on
\((\theta _1,\theta _2)\) , strictly increasing on \((0,\nu _0)\) , and strictly decreasing
on \((\nu _0,\infty )\) .

Since \(f\)  is positive on its support, then it must be increasing on \((0,\nu _0)\) . If \(f\) 
is concave on \((0,\theta _1)\) , then it is convex on \((\theta _1,\theta _2)\) , contradicting
the fact that \(\nu _0\in (\theta _1,\theta _2)\) . The convexity part of the lemma follows, and
since \(\theta _2>\nu _0\)  and \(f\)  is unimodal, then \(f\)  is decreasing on \((\nu _0,\infty )\) .
Moreover, since \(f^{\prime }>0\)  near 0 and \(f^{\prime }<0\)  near infinity, then \(f^{\prime }\)  has an odd number of
zeroes. Integrability of \(f\)  implies \(f^{\prime }\rightarrow 0\) , hence \(f^{\prime }\)  is non-zero everywhere on
\((0,\infty )\)  except at \(\nu _0\) , otherwise \(f\)  would have at least three inflection points.
Since \(f\)  is \(C^2\)  on \((0,\infty )\) , then (REF ) and (REF ) hold. Moreover,
\(h_p\)  is \(C^2\) . Let
\(*:=\lbrace p\in \lbrace 0\rbrace :\nu _p>(\nu _0+\theta _2)/2\rbrace . \footnote {\ If, as in Remark\ref {criteria relaxations}, we usethe condition \nu _1+c_1>\theta _2 rather than \nu _1>(\nu _0+\theta _2)/2, then instead let*:=\lbrace p\in \nu _p\ge \nu _1+c_1\rbrace . The arguments that follow still apply with trivialmodifications.}\) 
By assumption \(1\in *\) . We show that membership in \(*\)  is sufficient to make (REF )
a strict inequality.
Lemma 3.2 
\(h_p^{\prime }(c_p)>0\)  for all \(p\in *\) .

First we locate \(\nu _p+c_p\)  and \(\nu _p-c_p\) . Note that \(f(\nu _p-c)\)  is increasing and \(f(\nu _p+c)\) 
is decreasing for \(c\in (0,\nu _p-\nu _0)\) . The fact that \(h_p\)  is negative near 0 and \(h_p(c_p)=0\) 
implies \(c_p>\nu _p-\nu _0\) . By assumption \(\nu _p>(\nu _0+\theta _2)/2\) , so it follows that
\(\nu _p+c_p>\theta _2.\) 
On the other hand, we easily have \(\nu _p-c_p>0\) , otherwise \(h_p(c_p)=f(\nu _p+c_p)>0\) . If
\(\nu _p-c_p>\nu _0\) , then \(f(\nu _p-c_p)>f(\nu _p+c_p)\)  by Lemma REF , again contradicting
\(h_p(c_p)=0\) . Thus
\(0<\nu _p-c_p<\nu _0.\) 
Now suppose for the sake of contradiction that \(h^{\prime }_p(c_p)=0\) . Either \(\nu _p-c_p\in (0,\theta _1)\)  or
\(\nu _p-c_p\in [\theta _1,\nu _0)\)  by (REF ). If the latter holds, then \(f^{\prime \prime }(\nu _p-c_p)<0\) ,
and (REF ) implies \(f^{\prime \prime }(\nu _p+c_p)>0\) . Note that
\(h^{\prime \prime }_p(c_p)=f^{\prime \prime }(\nu _p+c_p)-f^{\prime \prime }(\nu _p-c_p)\) , hence \(h^{\prime \prime }_p(c_p)>0\) . By continuity of \(h^{\prime \prime }_p\)  and
(REF ), \(h_p\)  is strictly convex and thus positive in a neighborhood of \(c_p\)  (excluding
\(c_p\)  itself), contradicting the minimality of \(c_p\)  in \(\mathcal {S}_p\) .
Suppose instead that \(\nu _p-c_p\in (0,\theta _1)\) . Then from conditions (1) and (2) and equations
(REF ) and (REF ), we have the inequalities \(\nu _0f^{\prime }(\nu _p-c_p)>f(\nu _p-c_p)\)  and \(\nu _0f^{\prime }(\nu _p+c_p)>-f(\nu _p+c_p)\) . It follows that \(\nu _0h^{\prime }_p(c_p)>-h_p(c_p)=0\) 
and we obtain a contradiction.
Next, we prove some properties of \(c_p\) .
Lemma 3.3 
The map \(p\mapsto c_p\)  is continuously differentiable in \(*\) .

Define \(\psi :*\times \mathbb {R}_+\)  by \(\psi (p,c):=h_p(c)\) . By Proposition REF , \(\psi \)  is
differentiable and has partial derivatives
\(\frac{\partial }{\partial c}\psi (p,c)=f^{\prime }(\nu _p+c)+f^{\prime }(\nu _p-c)=h^{\prime }_p(c),\) 
\(\frac{\partial }{\partial p}\psi (p,c)=(f^{\prime }(\nu _p+c)-f^{\prime }(\nu _p-c))\frac{d\nu _p}{dp},\) 
both of which are jointly continuous in \(p\)  and \(c\) . Thus \(\psi \)  is continuously differentiable.
By Lemma REF , \(\frac{\partial }{\partial c}\psi (p,c_p)>0\)  for all \(p\in *\)  and so the
continuous differentiability of \(p\mapsto c_p\)  follows from (REF ) and the implicit
function theorem.
Lemma 3.4 
For any \(p\in *\) , \(\frac{dc_p}{dp}\)  and \(\frac{d\nu _p}{dp}\)  have the same sign, and
\(\big |\frac{dc_p}{dp}\big |>\big |\frac{d\nu _p}{dp}\big |\)  if they are non-zero.

By (REF ), we have \(f(\nu _p+c_p)-f(\nu _p-c_p)=0\) . The left side is differentiable in \(p\) 
by Lemma REF , so taking derivatives, we obtain
\(f^{\prime }(\nu _p+c_p)\left(\frac{d\nu _p}{dp}+\frac{dc_p}{dp}\right)- f^{\prime }(\nu _p-c_p)\left(\frac{d\nu _p}{dp}-\frac{dc_p}{dp}\right) = 0.\) 
Rearranging yields
\(\frac{dc_p}{dp} = \frac{d\nu _p}{dp}\left(\frac{f^{\prime }(\nu _p-c_p)-f^{\prime }(\nu _p+c_p)}{f^{\prime }(\nu _p-c_p)+f^{\prime }(\nu _p+c_p)}\right),\) 
where the fraction is well-defined with positive denominator by Lemma REF . The
numerator is positive by Lemma REF , (REF ), and (REF ), so
\(\frac{dc_p}{dp}\)  has the same sign as \(\frac{d\nu _p}{dp}\) . If \(\frac{d\nu _p}{dp},\frac{dc_p}{dp}>0\) , then one can see immediately from (REF ) that \(\frac{dc_p}{dp}>\frac{d\nu _p}{dp}\) .
The reverse also follows.
Our final lemma concerns a criterion for pointwise increasingness of \(\nu _p\) .
Lemma 3.5 
Fix \(p\in \lbrace 0\rbrace \) . If
\((\nu _p-c_p)\frac{f^{\prime }(\nu _p+c_p)}{f(\nu _p+c_p)} > -1,\) 
then \(\nu \)  is increasing at \(p\) .

Since \(f(\nu _p+c_p)=f(\nu _p-c_p)\) , we may rearrange to obtain
\(-\frac{f(\nu _p-c_p)}{\nu _p-c_p}<f^{\prime }(\nu _p+c_p).\) 
Define the line
\(\ell (x):=-\frac{xf(\nu _p-c_p)}{\nu _p-c_p}+\frac{2\nu _pf(\nu _p-c_p)}{\nu _p-c_p},\ x\ge \nu _p+c_p,\) 
and note that
\(\begin{split}\ell (\nu _p+c_p) &= f(\nu _p-c_p)=f(\nu _p+c_p), \\\ell (2\nu _p) &= 0=f(0).\end{split}\) 
Note that (REF ) implies \(f^{\prime }(\nu _p+c_p)=\ell ^{\prime }(\nu _p+c_p)\) , so by convexity of
\(f\)  on \((\theta _2,\infty )\) , we have \(f^{\prime }>\ell ^{\prime }\)  on \((\nu _p+c_p,\infty )\) . Via integration we obtain
\(f(\nu _p+c)>\ell (\nu _p+c)\) 
for \(c>c_p\) . We now split into two cases to show \(f(\nu _p+c)>f(\nu _p-c)\)  for \(c>c_p\) .
Case 1. Suppose \(\nu _p-c_p\in (0,\theta _1]\) . By (REF ) and the convexity of
\(f\)  on this interval, we have \(\ell (x)>f(2\nu _p-x)\)  for \(x\in (\nu _p+c_p,2\nu _p)\) , making use of
the fact that convexity is preserved under reflection and translation. A change of variables
gives the inequality \(\ell (\nu _p+c)>f(\nu _p-c)\)  for \(c\in (c_p,\nu _p)\) . Combining with
(REF ) yields \(f(\nu _p+c)>f(\nu _p-c)\)  for \(c\in (c_p,\nu _p)\) . If \(c\ge \nu _p\) ,
then \(f(\nu _p+c)>0=f(\nu _p-c)\)  and we are done.
Case 2. Suppose instead \(\nu _p-c_p\in (\theta _1,\nu _0)\) . If it happens that
\(\ell (\nu _p+c)>f(\nu _p-c)\)  for all \(c\in (c_p,\nu _p)\) , then the argument in Case 1 applies and we
are done. Otherwise, let
\(\tilde{\ell }(x):=-(x-\nu _p-c_p)f^{\prime }(\nu _p-c_p)+f(\nu _p+c_p),\ x\ge \nu _p+c_p,\) 
be the line such that
\(\tilde{\ell }(\nu _p+c_p)=f(\nu _p+c_p)\) 
and
\(\tilde{\ell }(\nu _p+c_p+f(\nu _p+c_p)/f^{\prime }(\nu _p-c_p))=0.\) 
Lemma REF  and convexity imply \(f^{\prime }(\nu _p+c)>\tilde{\ell }^{\prime }(\nu _p+c)\)  and thus
\(f(\nu _p+c)>\tilde{\ell }(\nu _p+c)\)  for all \(c>c_p\) . By concavity of \(f\)  near \(\nu _p-c_p\) , one
can easily see that \(\tilde{\ell }(\nu _p+c)>f(\nu _p-c)\)  for \(c\in (c_p,\nu _p-\theta _1]\) . It
remains to show \(\tilde{\ell }(\nu _p+c)>f(\nu _p-c)\)  for \(c\in (\nu _p-\theta _1,\nu _p)\) .
We have by assumption that \(f(\nu _p-c)>\ell (\nu _p+c)\)  for \(c\)  in a right neighborhood of \(c_p\) .
We also have by (REF ) that \(f(\nu _p-c_p)=\ell (\nu _p+c_p)\) . It follows that
\(-f^{\prime }(\nu _p-c_p)>\ell ^{\prime }(\nu _p+c_p)=-\frac{f(\nu _p-c_p)}{\nu _p-c_p}.\) 
Substituting with \(f(\nu _p-c_p)\)  with \(f(\nu _p+c_p)\)  and rearranging yields
\(\nu _p+c_p+\frac{f(\nu _p+c_p)}{f^{\prime }(\nu _p-c_p)}>2\nu _p.\) 
The left side is precisely the root of \(\tilde{\ell }\)  whereas the right side is a root of \(f\) .
We showed previously that \(\tilde{\ell }(2\nu _p-\theta _1)>f(\theta _1)\) . Convexity of \(f\)  on
\((0,\theta _1)\)  implies \(\tilde{\ell }(\nu _p+c)>f(\nu _p-c)\)  for \(c\in (\nu _p-\theta _1,\nu _p)\) ,
and we are done.
We have now shown that, in general, \(f(\nu _p+c)>f(\nu _p-c)\)  for \(c>c_p\) . We also know by
definition of \(c_p\)  that \(f(\nu _p+c)\le f(\nu _p-c)\)  for \(c<c_p\) . Thus the conditions of Lemma
REF  are satisfied and so \(\nu \)  is increasing at the point \(p\) .
To prove the main theorem, it suffices to show (REF ) for all \(p\in *\) 
and that \(*=\lbrace 0\rbrace \) . If we have conditions (1) and (2) as they are written, then (REF )
holds for all \(p\in *\)  immediately by (REF ) and (REF ). Suppose instead we only
have the weaker condition as stated in Remark REF (b):
\(\begin{split}f^{\prime }/f > 1/(\nu _1-c_1) &\text{ on } (0,\theta _1), \\f^{\prime }/f > -1/(\nu _1-c_1) &\text{ on } (\nu _1+c_1,\infty ).\end{split}\) 
Let \(u(p):=\nu _p-c_p\)  on \(*\) , so \(u\)  is differentiable on its domain. Also note that \(u>0\)  by
(REF ). Suppose for some \(p^{\prime }\in *\)  that \(u(p^{\prime })\le \nu _1-c_1\) . Then
\((\nu _{p^{\prime }}-c_{p^{\prime }})\frac{f^{\prime }(\nu _{p^{\prime }}+c_{p^{\prime }})}{f(\nu _{p^{\prime }}+c_{p^{\prime }})}>-1,\) 
i.e., (REF ) holds for \(p^{\prime }\)  and so \(\nu _p\)  is increasing at \(p^{\prime }\) . Lemma
REF  implies \(u\)  is decreasing at \(p^{\prime }\) . It follows that for all \(p\ge p^{\prime }\) , \(u\)  is
decreasing and so \(u(p)\le \nu _1-c_1\) . In particular, \(\nu _p\)  is increasing for all \(p\ge p^{\prime }\) .
By assumption, \(1\in *\)  and trivially \(u(1)\le \nu _1-c_1\) . Since \(*\in [1,\infty )\) , then
\(\nu _p\)  is increasing for all \(p\in *\) . Recall that \(p\in *\)  if \(\nu _p>(\nu _0+\theta _2)/2\) .
Since \(\nu _p\)  is increasing at \(p=1\)  and for all \(p\in *\) , then \(\nu _p>(\nu _0+\theta _2)/2\) 
for all \(p\in \lbrace 0\rbrace \) , hence \(\lbrace 0\rbrace =*\) . Clearly the argument
for \(*=\lbrace 0\rbrace \)  still applies if we use the alternative definition for \(*\) .
The proof directly extends to the case where \(f^{\prime \prime }\)  only has a single root \(\theta >\nu _0\)  by
setting \(\theta _1=0\) . In fact, conditions (1) and (2) are not necessary. Indeed, we use them
once in the proof of Lemma REF  in the case \(\nu _p-c_p\in (0,\theta _1)\) , but this
is no longer relevant if \(f^{\prime \prime }\)  has only one root. The only other time we use the conditions
is to prove that (REF ) holds for all \(p\in *\) . However, if \(f^{\prime \prime }\)  has only
one root then, (REF ) holds automatically. Indeed, note that \(f\)  is concave
and increasing near \(\nu _p-c_p\)  for all \(p\in *\) , hence
\(\frac{f(\nu _p-c_p)}{\nu _p-c_p}>f^{\prime }(\nu _p-c_p).\) 
Since \(f^{\prime }(\nu _p+c_p)<0\) , then by substituting \(f(\nu _p+c_p)=f(\nu _p-c_p)\) , we have
\((\nu _p-c_p)\frac{f^{\prime }(\nu _p+c_p)}{f(\nu _p+c_p)}>\frac{f^{\prime }(\nu _p+c_p)}{f^{\prime }(\nu _p-c_p)}.\) 
The right side dominates \(-1\)  as a consequence of Lemma REF , so we arrive at
(REF ). This proves the corollary.

Discussion
This work attempts to broaden our understanding of true skewness by demonstrating the true skewness of several additional distributions and by establishing simpler criteria for which one can conclude a distribution is truly skewed. Theorems REF –REF , which establish the parameter regions for which the Lévy, chi-squared, Weibull, and skew-normal distributions are truly positively skewed, all rely crucially on showing that the conditions of Lemma REF  hold for every relevant value of \(p\) . This is done by analyzing what we have called the “log density ratio of the left and right parts,” which for given \(p\)  is
\(R_p(x) = \log \left(\frac{f(\nu _p-x)}{f(\nu _p+x)}\right), \qquad x\in [0,\nu _p),\) 
where \(f\)  is the density function of the distribution in question. The conditions of Lemma REF  hold if one can show that \(R_p(\cdot )\)  has exactly one positive root \(x_0\) , satisfying \(R_p(x)>0\)  for \(0<x<x_0\)  and \(R_p(x)<0\)  for \(x>x_0\) . In Theorems REF , REF , and REF , this can be shown in quite a straightforward manner simply by finding the critical points of \(R_p(\cdot )\) . Such an approach may likely be applied to other visibly skewed distributions for which a closed-form expression for its density function exists.
However, for the skew-normal distribution, Theorem REF , the absence of a closed-form expression for its density function makes the computation of the critical points of \(R_p(\cdot )\)  intractible. Instead, we exploit the fact that the log derivative of a distribution function \(\Psi \)  is simply \(\psi /\Psi \) , where \(\psi \)  is the corresponding density function. The ratio \(\psi /\Psi \)  is the reciprocal of what is commonly referred to as the hazard rate of the distribution \(\Psi \) , which has been studied with some detail for most well-known distributions. In the proof of Theorem REF , we encounter the reciprocal of the hazard rate of the standard Gaussian, which goes by the special name of the Mills' ratio. The Mills' ratio, as one may expect, has been very well-studied, allowing us to employ a closed-form expression, which very tightly bounds the quantities we care about, to prove the required properties of the skew-normal distribution's log density ratio of the left and right parts.
Indeed, the proof of Theorem REF  identifies a key yet unsurprising connection between the notion of true skewness, which is fundamentally a comparison of the rate of decay of the two sides of a distribution, and the hazard rate. It suggests that the notion of true skewness is an accurate reflection, and in a much more rigorous fashion than Pearson's coefficients of skewness, of the essence of what we imagine skewness to mean. More practically, the method by which we prove Theorem REF  should work for a larger variety of distributions; a reasonable first direction would be skewed versions of other symmetric distributions, as introduced in, for example, Chapter 1 of [7]}.
The remainder of this section discusses sums and products of truly skewed random variables; examines whether true skewness extends to the discrete case; and provides an interpretation of true skewness for multivariate distributions.
Sums and products of truly skewed random variables
The goal of this paper and this discussion is to present tools for establishing a larger class of distributions which are truly skewed. A natural question is whether or not true skewness is preserved under certain “transformations” of truly skewed random variables.
Theorem REF  gives an affirmative answer to this question, but with limitations: transforming a random variable with a strictly decaying distribution by an increasing convex function preserves its true positive skewness. One might also expect that sums of truly positively skewed random variables preserve true positive skewness, but the rather strong requirement that the \(p\) -means of a truly positively skewed random variable be strictly increasing allows one to construct rather straightforward counterexamples. Indeed, we may even take the two summands to be identically distributed. For \(1/2<\lambda <1\) , let \(X\)  and \(Y\)  be i.i.d. with density
\(f(x):={\left\lbrace \begin{array}{ll}\lambda & 0 < x \le 1 \\1-\lambda & 1 \le x < 2 \\0 & \text{otherwise}.\end{array}\right.}\) 
Since \(f\)  is decreasing on its support, \(X\)  and \(Y\)  are truly positively skewed by Proposition REF . One can easily compute the density function \(f_Z\)  of \(Z=X+Y\) , the convolution of \(f\)  with itself and see that, depending on the value of \(\lambda \) , the \(p\) -means of \(Z\)  are not always increasing. For instance, for \(\lambda =3/5\) , computing numerically the median \(\nu _1^Z\)  and the integrals in (REF ) yields \(\nu _1^Z\approx 1.786\)  and
\(\int _0^{4-\nu _1}\log y\ f_Z(\nu _1+y)dy - \int _0^{\nu _1} \log y\ f_Z(\nu _1-y)dy\approx -0.000699 < 0,\) 
which implies that \(p\mapsto \nu _p^Z\)  is decreasing at \(p=1\) , and so \(Z\)  is not truly positively skewed. Numerical computations were carried out using Wolfram Mathematica, version 12.3.0.0.
On the other hand, it may be more fruitful to start with random variables with monotone density functions. One class of truly positively skewed densities that is closed under summation is the class of “decreasing linear densities,” of the form
\(f(x) = h - \frac{h^2x}{2}, \qquad x\in [0,2/h],\) 
for arbitrary \(h>0\) .

True skewness of discrete distributions
In the discrete case, even the simplest sums of truly skewed random variables fail to remain truly skewed. Take \(X\)  and \(Y\)  to be independent \(\operatorname{Bernoulli}(1/3)\)  random variables; one can show that these are truly positively skewed quite easily. Their sum is \(\operatorname{Binomial}(2,1/3)\) , which has median 1 and mean 2/3, hence \(\nu _2<\nu _1\) , and true positive skewness fails to hold. Indeed, the notion of true skewness is finnicky for discrete distributions, at least in the way it has been formulated here and in [8]}: this could be due to the interpretation of discrete distributions as limits of successfully sharper (continuous) bump functions, in which case distributions like the \(\operatorname{Binomial}(n,1/3)\)  are no longer unimodal but in fact have \(n\)  modes.
The proof techniques used in this paper often involve taking the log of a ratio of the density function
of a distribution. In particular, the logarithm of a product of two entities is equal to the sum of the individual
logarithms of those entities. For this reason, it seems natural to examine how true skewness behaves under
the products of random variables and whether it is preserved. In particular, it seems that taking the logarithm
of the density of product of two random variables would yield a sum which could then imply true skewness
depending on how one defines the two random variables.

True skewness of continuous multivariate distributions
The definition of Fréchet \(p\) -mean (Def. REF ) extends naturally to multivariate distributions as follows.
Let \(\mathbf {X} \in \mathbb {R}^k\)  be a random vector. The \(p\) -mean \(\nu _p = \left(\nu _p^{(1)}, \nu _p^{(2)}, \ldots , \nu _p^{(k)}\right)\)  of \(\mathbf {X}\)  is defined
\(\nu _p = \operatornamewithlimits{arg\,min}_{\mathbf {a} \in \mathbb {R}^k} E \big [\Vert \mathbf {X} - \mathbf {a}\Vert ^{p}\big ],\) 
where \(\Vert \cdot \Vert \)  is the usual Euclidean norm.
In the univariate setting, true skewness corresponds to the sign of \(d\nu _p/dp\)  representing the direction of trajectory \(\nu _p\) .
Following [8]}, we adjust true skewness accordingly. We let
\(\tau _p :=\frac{d\nu _p}{dp} \Big / \left\Vert  \frac{d\nu _p}{dp} \right\Vert \) 
denote the unit tangent vector for trajectory of \(\nu _p\)  in \(\mathbb {R}^k\) . We will illustrate true skewness in a multivariate setting by means of an example.
It was conjectured in [8]} that the limiting direction vector \(\lim \limits _{p \rightarrow \sup \mathcal {D}} \tau _p\)  with \(p\)  increasing to the rightmost bound in its domain \(\mathcal {D}\)  may be interpreted as the “direction” of true skewness.
As an example, considered a multivariate skew normal distribution defined in [11]}, the probability density function of a multivariate skew-normal random vector \(\mathbf {Y} \in \mathbb {R}^k\)  is
\(f(\mathbf {y}) = 2 \phi _k(\mathbf {y}; \mu , \Sigma ) \Phi _1(\lambda ^\top \Sigma ^{-1/2}(\mathbf {y} - \mu )) \qquad \mathbf {y} \in \mathbb {R}^k\) 
where \(\phi _k(\cdot \,;\mu , \Sigma )\)  is the density function of the \(k\) -variate normal distribution with mean \(\mu \)  and covariance matrix \(\Sigma \) , and \(\Phi _1(\cdot )\)  is the cumulative distribution function of the univariate standard normal distribution. Taking \(\lambda  = 0\)  recovers the standard multivariate normal distribution with density \(\phi _k(\cdot ; \mu , \Sigma )\)  . We refer to \(\lambda \)  as the skewness parameter vector.
<FIGURE>The arrows plotted in Figure REF  motivate the interpretation of multivariate true skewness via direction vectors \(\tau _p\) , which are naturally co-linear with the skewness parameter vector \(\lambda \) .

Acknowledgments
The majority of this research was done as a part of a Research Experience for Undergraduates (REU) program at the Oregon State University Department of Mathematics funded by NSF grant DMS-1757995. We would like to thank Javier Rojo of Indiana University for his helpful feedback on the concepts considered in this paper. We would also like to acknowledge professor Holly Swisher of Oregon State University who as the director of the REU program provided invaluable perspective throughout the summer.
